Universal gravitation and inclines

There's an inclined plane with theta unknown. The frictional coefficient is 0. m1 is higher on the inclined plane than m2.

m1 = 1680kg
m2 = 152kg
Distance between the two: 11mm

At what angle of inclination will the 2nd mass begin to slide down the plane?

Normally (without 2 objects) I know that net force would have to equal 0 in order for the box to slide down. In other words, it would be Fgx - Ff = Fnet = 0.

I first started with this:

1489.6sinθ - 0 = 0 But I know the law of universal gravitation plays a part in this. I was thinking about making the 1489.6sinθ equal to the universal gravitational equation since I have all the variables.

1489.6sinθ = (Gm1m2)/(r^2)

Would this be the correct thing to do? If so, I'm confused as to why they would be equal. Thanks.

If your physics class is calculus based, the 2nd attachment in the link below illustrates a simple, structured methodology for approaching problems like this - it even has a mass on incline example. Check it out.

Oh no, I know how to do this type of a problem when there is a single mass. But this question is implying that m1 is exerting a gravitational force on m2 and vice versa. (How do I know this for sure? The question provides a given: G=6.67259e-11, BIG hint) That's what I'm confused about.

Ok basicly, you do a sum of forces like you did, the force which will counterbalance the Weight of mass 2 will be the gravitational force mass 1 exerts on mass 2.
[tex] \sum_{i=1}^{n} \vec{F}_{i} = \vec{Fg}_{12} + m_{2} \vec{g} = \vec{0} [/tex]